# Mathematics Paper 2 Questions and Answers - Momaliche Pre Mock Exams 2023

INSTRUCTIONS TO CANDIDATES

• This paper contains TWO sections: section I and section II
• Answer ALL the questions in Section I and only five questions from section II.
• Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.
• Marks may be given for correct working even if the answer is wrong.
• Non-programmable silent electronic calculators and KNEC mathematical tables may be used except where stated otherwise.

SECTION 1 (50 MARKS)

Answer all questions in this section in the spaces provided.

1. A positive two digit number is such that the product of the digits is 24. When the digits are reversed, the number formed is greater than the original number by 18. Find the number.          (3mks)
2. Use tables of squares, square roots and reciprocals to evaluate                       (4mks)
234     +        16
√0.02698     (0.18149)2
3. The height and radius of a cone are measured as 21 cm and 14.0 cm respectively. Taking π= 3.142, find the percentage error in the volume of the cone.                               (3mks)
4. Express the following in surd form and simplify by rationalizing the denominator without using a calculator and leave your answer in the form  a + b√c       (3mks)
1 + Cos 30°
1− Sin 60°
5. Solve for x in : Log2(x + 7) – Log2(x – 7) = 8                                                        (3mks)
6. A businessman obtained a loan of Ksh 450,000 from a bank to buy a Matatu that was valued at the same amount. The bank charges interest at 24% per annum compounded quarterly per year. Calculate the total amount of money the businessman paid to clear the loan in 4½ years to the nearest shilling.       (3mks)
7. In the diagram below, BT is a tangent to the circle at B. AXCT and BXD are straight lines. AX = 6cm, CT = 8cm, BX = 4.8cm and XD = 5cm. Find the length of BT.    (3Marks)
8. Find the possible values of x given that is a singular matrix.        (3mks)
9. The cost C of operating an electronic business is partly constant and partly varies as the square of labour input L. If C=25,000 when L=20 and C=45,000 when L=30. Find C when L=8.    (3Mks)
10. The 2nd, 4thand 7th terms of an A.P. are the first 3 consecutive terms of a G.P. Find the common ratio of the G.P if the common difference of the A.P. is 2.      (3mks)
11. P and Q are two points such that OP=i + 2j + 3k and OQ = 4i + 5j – 3k. M is a point that divides PQ externally in the ratio 3:2. Find the co-ordinates of M, given that O is the origin.                     (3mks)
12. A circle Centre C (5, 5) passes through points A (1, 3) and B (a, 9). Find the equation of the circle and hence the possible values of a.                              (4mks)
13. Tap A can fill an empty tank in 3 hours, while tap B can fill the same tank in 2 hours. When the tank is full, tap C can empty the tank in 5 hours. Tap A and C are opened for 4 hours and then closed.
1. Determine the fraction of the tank that is still empty.                          (1mks)
2. Find how long it would take to fill the remaining fraction of the tank if all the three taps are opened.     (2mks)
14. Determine the interquartile range for the following set of numbers.                         (2mks)
4, 9, 5, 4, 7, 6, 2, 1, 6, 7, 8.
15. Solve the equation Sin(3x – 10)° =  0.4337 for 0°≤ Ɵ ≤180°                               (3mks)
16.
1. Expand and simplify (3x – y)4(2mks)
2. Use the first three term of the expansion to approximate the value of (6 – 0.2)4.  (2mks)

SECTION II (50MARKS) ANSWER ANY 5 QUESTIIONS ONLY

1. Mrs. Mutua earns a basic salary of K£ 12,000 p.a. and is housed by the employer at a nominal rent of Shs 1,200 per month. She is entitled to a personal relief of K£ 1,320 p.a. and a premium relief of 10% on her insurance premium of K£ 800 p.a. The table of tax rate is as below.
 Taxable income (K£ p.a.) Rate (%) 1 – 2100 10 2101 – 4200 15 4201 – 6300 20 6301 – 8400 25 Over     8400 30
Calculate;
1. Calculate the net tax per annum.                                (7mks)
2. Other deductions includes W.C.P.S  Shs 600 per month, NHIF Shs. 500 per month. Calculate her net pay per month.   (3mks)
2. The Line AB = 5cm is a side of a triangle ABC in which angle ABC = 90⁰ and angle BAC = 60⁰.
1. Construct triangle ABC (2mks)
2. Construct the Locus P such that angle APB = angle ACB (2mks)
3. Locate by construction points Q1 and Q2 which satisfy the conditions below:
1. Q1 and Q2 lie on the same side of line AB and C(3mks
2. Area of triangle AQ1B = Area of triangle AQ2B = ¾Area of triangle ABC
3. Angle AQ1B = Angle AQ2B = 30⁰
Measure the length of the line Q1Q2 (3mks)
4. Calculate the area above the line Q1Q2 bounded by the locus of point P       (3mks)
3. The diagram below shows a square based pyramid V vertically above the middle of the base.  PQ = 10cm and VR = 13cm.  M is the midpoint of VR. Find to 2 decimal places
1. the length PR.   (2mks)
2. The height of the pyramid.      (2mks)
1.
1. the angle between VR and the base PQRS.    (2mks)
2. The angle between MR and the base PQRS.       (2mks)
3. The angle between the planes QVR and PQRS.                                (2mks)
4.
1. Complete the table below for y = sin 2x and y = sin ( 2x + 30) giving values to 2d.p(2mks)
 X 0 15 30 45 60 75 90 105 120 135 150 165 180 Sin 2x 0 0.87 −0.87 0 Sin ( 2x +30) 0.5 0.5 −1 0.5
2. Draw the graphs of y=sin 2x and y = sin (2x + 30) on the axis.(4mks)
3. Use the graph to solve [Image_Link]blob: sin (2x + 30) − sin 2x = 0. (1mk)
4. Determine the transformation which maps [Image_Link]blob:sin 2x onto sin (2x + 30).  (1mk)
5. State the period and amplitude ofy = sin (2x + 30)     (2mks)
5. OABC is a parallelogram with verities 0(0,0), A(2,0)  B(3,2) and C(1,2). O,A,B,C   is the image of OABC under transformation matrix. 1.
1. Find the coordinates of O1A1B1C1.    (2mks)
2. On the grid provided, draw OABC and O1A1B1C1    (2mks)
2.
1. Find O11A11B11C11, the image of O1A1B1C1 under transformation matrix (2mks)
2. On the same grid draw O11A11B11C 11 (1mk)
3. Find a single matrix that maps O11A11B11C11 onto OABC               (3mks)
6. The following table shows the distribution of marks obtained by 50 students in a test.
 Marks 45-49 50-54 55-59 60-64 65-69 70-74 75-79 No. of Students 3 9 13 15 5 4 1
By using an assumed mean of 62, calculate
1. The mean.  (5mks)
2. The variance. (3mks)
3. The standard deviation.   (2mks)
7. A box contains 3 brown, 9 pink and 15 white cloth pegs.  The pegs are identical except for the colour.
1. Find the probability of picking.
1. A brown peg.   (1mark)
2. A pink or a white peg.  (2 marks)
2. Two pegs are picked at random, one at a time without replacement.  Find the probability that:
1. Atleast one brown peg is picked.   (4marks)
2. both pegs are of the same colour.   (3marks)
8. A wholesaler stocks two types of rice: Refu and Tamu. The wholesale prices of 1 kg of Refu and 1 kg of Tamu are Ksh 80 and Ksh 140 respectively. The wholesaler also stocks blend A rice which is a mixture of Refu and Tamu rice mixed in the ratio 3 : 2.
1.
1. A retailer bought 10 kg of blend A rice. To this blend, the retailer added some Tamu rice to prepare a new mixture blend X. The .ratio of Refu rice to Tamu rice in blend X was 1:2.
Determine the amount of Tamu rice that was added.   (3marks)
2. The retailer sold blend X rice making a profit of 20%. Determine the selling price of 1 kg of blend X.  (3 marks)
2. The wholesaler prepared another mixture, blend B, by mixing x kg of blend A rice with y kg of Tamu rice. Blend B has a wholesale price of Ksh130 per kg.Determine the ratio x : y.         (4mks)

### MARKING SCHEME

 1 Xy = 24X=24/y……………….…..(i)(10y + x) – (10x + y)= 189y – 9x = 18Y – x = 2 ……………………(ii)M1SubstitutingY-24/y= 2Y2 -2y – 24 =0Y2- 6y + 4y -24 =0M1(solving quadratic equation)Y(y-6) + 4(y-6) =0(y+4)(y-6)=0Ignoring negative value y=6    while x=4 The number is 46       A1 Total  3mks 2 234         +           16          √(2.698×10−1       (1.8149 × 10−1)2       234           +           16          1.6426 × 10−1        3.294 ×10−2234(0.6087 × 10) + 16(0.3036 × 102)234 × 6.087 + 16 × 30.36= 1910.118 Total 4mks 3 Actual volume        = 1/3 x 3.142 x 212 x 14.0     =  6466.236 Minimum volume   =1/3 x 3.142 x 20.52x 13.95  =  6139.9786    Maximum volume  = 1/3 x 3.142 x 21.52 x 14.05 = 6802.024         M1 Absolute error = max − min = 6802.024 − 6139.97862 = 331.0227                   M1                                  2                             2% error  = Absolute error  x100 = 331.0227 x 100= 5.1192%                 M1                 actual volume               6466.236 Total3mks 4 Cos 30° =Sin 60° = √3/2                               M1(for correct sine and cos) (1 + √3/2 )( 1 + √3/2) =  1 + √3/2 + √3/2 +¾(1 − √3/2)(1 + √3/2 )           1 − ¾                  M1(for correct rationilization)7/4 + 2√3/2                          ¼7/4+ 4√3                        A1 Total 3 Mks
 5 Log2x+7 =Log28           M1(for correct Log28)        x-7Dropping the logsx+7 = 8                                                        B1(for dropping logs)x-7x + 7 = 8x – 5663 = 7xX = 9                                                           A1 Total 3 mks 6 A = 450,000(1 + 24/100 x ¼)18               M(for correct substitution)A= 1,284,453                                                       A1 Total 2 mks 7 AX.CX = DX.XB6XC = 5 × 4.8    M1XC = 5 × 4.8 = 4 cm r² = (x − 9)²2 + (y − b)²CA = √((5−1)² + (5−3)²)     = √(4² + 2²)    = √20 units (x−5)² + (y−5)² = (√20)² x² − 10x + 25 + y² − 10y + 25 = 20 x² + y² − 10x − 10y + 50 = 20 x² + y² − 10x − 10y + 30 = 0 a² + 9² − 10a − 90 + 30 = 0 a² + 81 − 10a − 60 = 0 a² − 10a + 21 = 0 a = −b ± √(b² − 4ac)                  2a   = 10 ±√(100−(4×1×21))                  2  = 10 ± 4         2 ∴ a = 7 or 3 Total 4mks 13 A can fill  1/3 of the tank in 1 hr.C can empty  1/5 of the tank in 1 hr.Fraction filled by both tanks in 1 hr =  1/3 − 1/5 =  2/15Fraction filled in 4 hrs =  2/15 x 4 =  8/15Empty fraction = 1 −  8/15 = 7/15[Image_Link]blob:   A1 Fraction filled in one hour by all pipes =  1/3 + 1/2 − 1/5 = 19/30[Image_Link]blob:  M1Time to fill the empty fraction = 7/15 ÷  19/30 =  7/15 x 30/19 =  14/19hours[Image_Link]blob:  or=0.7368 hours  A1 Total 4mks
 14 Arrange numbers in ascending order1, 2, 4, 4, 5, 6, 6, 7, 7, 8, 9.Lower quartile Q1       =  4  M1Upper quartile Q3       = 7interquartile range   = 7 – 4 = 3  A1 Total 2mks 15 (3x – 10)° = Sin-10.4337 = 25.7°[Image_Link]blob:   M1(3x – 10)° = 25.7°, 154.3°, 385.7°, 514.3°, 3x= 35.7°, 164.3°, 395.7°, 524.3°X=  35.7,   164.3,  395.7,  524.3    M1         3          3          3          3X= 11.9°,  54.77°,  131.9°,  174.77°     A1 (Award 1 mark if all the angles are there). Total 3mks 16 (3x)4 – 4(3x)3(y) + 6(3x)2(y)2 – 4(3x)(y)3 + (y)4[Image_Link]blob:      M181x4 – 108x3y + 54x2y2 – 12xy3 + y4       A1 3x=6     X=2  and y=0.281(24) – 108(23)(0.2) + 54(22)(0.22)[Image_Link]blob:     M1(for correct substitution)1286 – 172.8 + 8.64= 1131.84[Image_Link]blob:     A1 17 (a) Net Tax per annumTaxable income p.a = ( K£12,000 + 15% of K£12,000 −  1.200 x 12)                                                                                                  20                                = K£ 13,080 p.a.[Image_Link]blob:    M11st band 2100 x 10% = 2102nd band 2100 x 15% = 315 B13rd band 2100 x 20%  = 420[Image_Link]blob: B14th band 2100 x 25% =  525[Image_Link]blob: B1Remaining 4680 x 30%   = 1,404[Image_Link]blob: B1Gross tax      = K£ 2874 P.a.[Image_Link]blob: M1Net tax         = Gross tax – reliefNet tax         = K£ (2874 – 1320 – 80)                     = K£ 1474 p.a.[Image_Link]blob: A1 (b) Deductions per monthTax per month =  1474 x 20 = 2,456. 67                                   12Premium per month     =  800 x 20 = Shs. 1,333.33                                             12Total deductions per month = 2,456. 67 + 1,333.33 + 600 + 500 = Shs 4890 p.m.[Image_Link]blob: M1Taxable income per month =  13080 x 20 = 21,800[Image_Link]blob:   M1                                                       12Net pay = Ksh 21, 800 – 4890 = Shs 16, 910[Image_Link]blob:.    A1 Total 10 mks
 18. Q1Q2 = 9.0cm B1    ±0.1cm Area of segment = Area of sector OQ1Q2 − Area of ΔQ1Q2                           = 129/360 × π × 5 × 5 − ½ × 5 × 5 sin 129°  M2 measure/Q1OQ2 in the diagram                          = 28.14 − 9.714                           = 18.416cm2 A1 Total = 10 mks 19. PR  = √(102 + 102) = √200  M1=  14.14   [Image_Link]blob:1 VO2  =  132  − 7.072  =  169 – 49.98 = √119.02  M1VO    =  √119.02= 10.91 A1   Tan [Image_Link]blob:θ = 10.91 M1              7.07θ = Tan -1 10.91 = 57.06°    A1                  7.07 Angle between MR and base PQRS =  Tan -1 10.91 M1                                                                          7.07                                                           = 57.06°     A1 Tan ᾱ  =  10.91. M1                   5            = 65.36°.   A1

1.
1. Complete the table below for y=sin 2x and y=sin ( 2x + 30) giving values to 2d.p.(2mks)
 x 0 15 30 45 60 75 90 105 120 135 150 165 180 Sin 2x 0 0.5 0.87 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0 Sin (2x+30) 0.5 0.87 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0 0.5
2. Draw the graphs of y=sin 2x and y = sin (2x + 30) on the axis.     (4mks) 3. Use the graph to solve [Image_Link]blob:Sin (2x + 30) − sin 2x = 0  (1mk)
x = 37.5° or 125.5°
4. Determine the transformation which maps  sin 2x onto sin (2x + 30)  (1mk)
Translation 5. State the period and amplitude of [Image_Link]blob:y = sin (2x + 30). (2mks)
Period 180° amplitude 1unit

2.  3. Let A = 62
 Marks f x D=x-A fd d² fd² 45-49 3 47 -15 -45 225 675 50-54 9 52 -10 -90 100 900 55-59 13 57 -5 -65 25 325 60-64 15 62 0 0 0 0 65-69 5 67 25 25 25 125 70-74 4 72 40 40 100 400 75-79 1 77 15 15 225 225 F = 50 Efd - 120 =2650
B1 d value
B1 fd2 ratio
1. Mean x = A +  Efd
Ef
62 + -120 / 50 M1
= 62 – 2.4  =  59.6  A1
2. v = M1
= 2560 − (120)2 = 53 − 5.76   M1A1
50       50
3. s.d = = √47.24
= 6.873. A1
4.
1.
1. P(B)
= 3/27 = 1/9
2. P(P) or P(W)
9/27 + 15/27 = 24/27 = 8/9
2. 1. P(BB) = (3/27 × 2/26) + (3/27 × 9/26) + (3/27 × 15/26) + (9/27 × 3/26) + (15/27 × 3/26)
= 1/117 + 1/26 + 5/78 + 1/26 + 5/78 = 25/117
2. P(BB) or P(PP) or P(WW)
= (3/27 × 2/26) + (9/27 × 8/26) + (15/27 × 14/26)
= 1/117 + 4/39 + 35/117
= 16/39
5.
1.
1. R:T
3:2
Cost of 1kg of the blend A = (3 × 80) + (2 × 140)
3 + 2
= 240 + 280
5
= Ksh. 104
R → 3/5 × 10 = 6kgs
T → 2/5 × 10 = 4kgs
Blend X → R:T
1:2
6     = ½
4 + x
4 + x = 12
x = 8kgs
2. Cost of 1kg of the blend A = (6 × 80) + (12 × 140)
18
= Ksh. 120
If 100% = Ksh. 120
∴ 120% = (120 × 120)
100
= Ksh. 144
2. Cost of 1kg of the blend B = 104x + 140y
x + y
∴     130     =  104x + 140y
x + y
130x + 130y = 104x + 140y
26x = 10y
13x = 5y
x/y = 5/13
x : y
5 : 13

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